Help with proving that the torsion subgroup of $y^2=x^3+x$ is $E(\mathbb{Q})_{tors} \cong \mathbb{Z}/2\mathbb{Z}$

Question

Let $E: y^2= x^3 + x$ be an elliptic curve over $\mathbb{Q}$. I'm trying to prove that $E(\mathbb{Q})_{tors} \cong \mathbb{Z}/2\mathbb{Z}$. In order to do that, I've already shown that $|E(\mathbb{F}_p)| \equiv 0 \mod 4$ for every prime $p \geq 3$, but I'm lost on what I'm supposed to do next.

Any help would be dearly appreciated.

EDIT: I do know the Nagell-Lutz theorem but I don't think I'm allowed to use it (Nagell-Lutz is described in the following section of the notes, so this exercise shouldn't need it). I've looked at the injection map described by Álvaro, and now I've successfully computed that the order of the torsion group is either $1,2$ or $4$. Order $1$ is no good, as $(0,0)$ has order $2$. How do I rule out a torsion group of order $4$?

EDIT $2$: I've found a hint online which states that $2P=(0,0)$ (and so has order $4$) iff $1=4d^4$ for some $d$. This is impossible; so the torsion group has to have order $2$. However, they don't give any explanation where this result comes from. Any ideas?

ATTEMPT: Is this correct? The point $(0,0)$ has order $2$, because the $y$-coordinate is zero. If $-1$ is a square, say $-1 = d^2$, then the equation takes the form $y^2 = x(x-d)(x+d)$ and there are three points of order two in $E(\mathbb{F}_p)$. Hence, if the index of $E(\mathbb{Q})$ is two (when it solely consists of $2$-torsion points), it is isomorphic $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ when $-1$ is a square and to $\mathbb{Z}/2\mathbb{Z}$ otherwise. Because $-1$ is a not a square in $\mathbb{Q}$, the result follows.

Answer 1

I'm pretty much ignorant about the higher powered theory of elliptic curves, but given that you know $|E(\Bbb{Q})_{tors}|$ to be a factor of $4$, the problem is reduced to a simple calculation.

We have $$E(\Bbb{C})[2]=\{P_\infty,(0,0),(i,0),(-i,0)\}.$$ So if your torsion group has order four, it cannot be the Klein 4-group. Consequently it has to be cyclic of order four. Therefore there must exist a point $P=(a,b)\in E(\Bbb{Q})$ such that $2P=(0,0)$. Because $(0,0)$ is its own negative, this implies that the tangent $T$ through $P$ also passes through the origin. The slope of this tangent is thus $$ \frac ba=\frac{3x^2+1}{2y}=\frac{3a^2+1}{2b} $$ implying that $$ 2b^2=3a^3+a. $$ But $P\in E$, so we also have $$ b^2=a^3+a. $$ Eliminating $b$ from this pair of equations gives $a^3=a$, and I'm sure you can take it from there.

Answer 2

Here is an alternative approach. Suppose $E(\mathbb{Q})$ contains a point $P$ of order $4$. Since there is a single point of order $2$ over $\mathbb{Q}$, namely $Q=(0,0)$, we must have $2P=Q=(0,0)$. Now we can use the formulas for multiplication by $2$ (Silverman's "Arithmetic of Elliptic Curves", page 54): $$x([2]P) = \frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6},$$ where, in your case, $$b_2=0,\ b_4= 2,\ b_6= 0,\ b_8= -1$$ so $$x([2]P) = \frac{x^4-2x^2+1}{4x^3+4x}.$$ Since $2P=Q=(0,0)$, the $x$-coordinate of $P$ satisfies $$x^4-2x^2+1=0$$ or, equivalently, $(x^2-1)^2=0$. Thus, $x=\pm 1$ and so, the $y$-coordinate of $P$ satisfies $y^2=2$ or $y^2=-2$, and so $y=\pm \sqrt{\pm 2}$. Since none of these $y$-coordinates are in $\mathbb{Q}$, we conclude that $P$ cannot be in $E(\mathbb{Q})$ to begin with.